A) \[\frac{1}{4}m{{\omega }^{2}}{{A}^{2}}\]
B) \[\frac{1}{4}{{m}^{2}}{{\omega }^{2}}{{A}^{2}}\]
C) \[{{\omega }^{2}}{{A}^{2}}\]
D) zero
Correct Answer: A
Solution :
Average energy \[=\frac{\int_{0}^{T}{Udt}}{\int_{0}^{T}{dt}}=\frac{1}{T}\int_{0}^{T}{Udt}\] \[=\frac{1}{2T}\int_{0}^{T}{m{{\omega }^{2}}{{A}^{2}}\,{{\cos }^{2}}\,(\omega t+\phi )dt}\] \[=\frac{1}{4}m{{\omega }^{2}}{{A}^{2}}\]
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